Collision Problem

1. Mar 2, 2008

the7joker7

1. The problem statement, all variables and given/known data

1) A 10.0-g object moving to the right at 20 cm/s makes an elastic head-on collision with a 15.0-g object moving in the opposite direction at 30.0 cm/s. Find the velocity of each object after collision.

2. Relevant equations

m$$_{1}$$v$$_{i1}$$ + m$$_{2}$$v$$_{i2}$$ = m$$_{1}$$v$$_{f1}$$ + m$$_{2}$$v$$_{f2}$$

(.5)m$$_{1}$$v$$_{1i}$$$$^{2}$$ + .(5)m$$_{2}$$v$$_{2i}$$$$^{2}$$ = (.5)m$$_{1}$$v$$_{1f}$$$$^{2}$$ + .(5)m$$_{2}$$v$$_{2f}$$$$^{2}$$

3. The attempt at a solution

I plugged values into the equations and got...

(.01)(.2) + (.015)(-.3) = (.01)(v$$_{f1}$$) + (.015)(v$$_{f2}$$)

(.5)(.01)(.2^2) + (.5)(.015)(-.3^2) = (.5)(.01)(v$$_{f1}$$$$^{2}$$) + (.5)(.015)(v$$_{f2}$$$$^{2}$$)

But I'm still stuck with two variables. How do I solve one of them?

2. Mar 2, 2008

hunter151

What is conserved in an elastic collision? (momentum and.........)

3. Mar 2, 2008

the7joker7

Energy.

I was under the impression I had the equation for conservation of energy there (the 2nd one).

4. Mar 2, 2008

hunter151

Oh wow, Im an idiot. You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.

For example, you could find Vf1 in terms of Vf2 from the first equation, and plug this Vf1 into the second equation. Now you are faced with an equation of just Vf2 as unknown.

5. Mar 2, 2008

the7joker7

Thanks.

I'm trying to do that now, and...

.0002 - .000675 = (.005)vf1^2 + (.0075)vf2^2

.005vf1^2 = -.000475 - .0075vf2^2

vf1^2 = .095 - 1.5vf2^2

vf1 = .3082 - 1.5vf2

-.0025 = (.01)(.3082 - 1.5vf2) + (.015)(vf2)

-.005582 = -(.015vf2) + (.015vf2)

Well, my vf2s cancel out :/

Did I make a math error?

6. Mar 2, 2008

hunter151

When you took the square root of Vf1, you forgot that there are two solutions, + and -. ;)